Complex Numbers making Real Numbers

As a point of curiosity, I found myself wondering when a complex number to an integer power creates a real number. For the sake of completeness, I also looked at when the result is a fully imaginary number.

More rigorously, define $\mathbb{C}^*$ as the set of complex numbers $a + bi$ where both $a$ and $b$ are non-zero, and $k\in\mathbb{N}^*$. For what values $x \in \mathbb{C}^*$ is $x^k \in \mathbb{R} \lor \mathbb{I}$?

k = 1

This is the trivial case. Since $x^1 = x$, there are no real or imaginary cases.

k = 2

$(a + bi)^2 = a^2 + 2abi + b^2i^2 = a^2 + 2abi – b^2 = (a^2 – b^2) + 2abi$.

This is real when the coefficient of $i$, i.e., $2ab$ is equal to 0. This is only true when $a = 0$ or $b = 0$, so it is not true for any element of  $\mathbb{C}^*$.

This is imaginary when $a^2 – b^2 = 0 \Rightarrow a^2 = b^2$. That is, $(a \pm ai)^2 \in \mathbb{C}^*$.

k = 3

$(a + bi)^3 = a^2 + 3a^2bi – 3ab^2 – b^3i = a(a^2 – 3b^2) + b(3a^2 – b^2)i$.

This is real when $b(3a^2 – b^2) = 0$. If $b = 0, x\notin\mathbb{C}^*$, so we ignore that case. $3a^2 – b^2 = 0 \Rightarrow 3a^2 = b^2$, so $b = \pm\sqrt{3} a$. Hence, if $x\in\mathbb{C}^*$, $x = a \pm a\sqrt{3}i$.

Similarly, this is imaginary when $a(a^2 – 3b^2) = 0$, so $b = \pm\frac{\sqrt{3}}{3} a$ and $x = a \pm\frac{a\sqrt{3}}{3}i$.

k = 4

Using similar methods, we get: $(a + bi)^4 = (a^4 – 6a^2b^2 + b^4) + 4ab(a^2 – b^2)i$.

This is real when $a^2 = b^2$, so $x = a \pm ai$.

Replacing $a^2 = c$ and $b^2 = d$ gives us the real part of $d^2 – 6dc + c^2$. Applying the quadratic formula with $a = 1$, $b = -6c$, and $d = c^2$ yields $c = \frac{6c \pm \sqrt{36c^2 – 4c^2}}{2} = 3c \pm c\sqrt{8} = (3 \pm 2\sqrt{2}) c$. Note that since the quadratic is symmetrical, that means that $3 + 2\sqrt{2} = \frac{1}{3 – 2\sqrt{2}}$.

Since $a^2 = c$ and $b^2 = d = (3 \pm 2\sqrt{2})c$, $a = \pm\sqrt{c}$ and $b = \pm(3 \pm 2\sqrt{2})\sqrt{c} = \pm(3 \pm 2\sqrt{2})a$.

This is imaginary, then, when $x = a \pm(3 \pm 2\sqrt{2}) ai$.

k = 5

This time, we get: $(a + bi)^5 = a(a^4 – 10a^2b^2 + 5b^4) + b(5a^4 – 10a^2b^2 + b^4)i$

This is real when $5a^4 – 10a^2b^2 + b^4 = 5c^2 – 10cd + d^2 = 0$.

$d = \frac{10c \pm\sqrt{100c^2 – 20c^2}}{10} = \frac{10c \pm 4c\sqrt{5}}{2} = 5 \pm 2\sqrt{5} c$, so $b = \pm\sqrt{5 \pm 2\sqrt{5}}a$.

This is real, then, when $x = a \pm \sqrt{5 \pm 2\sqrt{5}}ai$.

This is imaginary when $a^4 – 10a^2b^2 + 5b^4 = c^2 – 10cd + 5d^2 = 0$.

$d = \frac{10c \pm \sqrt{100c^2 – 20c^2}}{10} = \frac{10c \pm 4c\sqrt{5}}{10} = (1 \pm \frac{2\sqrt{5}}{5})c$, so $b = \pm\sqrt{1\pm\frac{2\sqrt{5}}{5}}a$.

This is imaginary, then, when $x = a \pm\sqrt{1\pm\frac{2\sqrt{5}}{5}}ai$.

k = 6

This time, we get: $(a + bi)^6 = (a^6 – 15a^4b^2 + 15a^2b^4 – b^6) + 2ab(3a^4 – 10a^2b^2 + 3b^4)i$

This is real when $3a^4 – 10a^2b^2 + 3b^4 = 3c^2 – 10cd + 3d^2 = 0$.

$d = \frac{10c \pm \sqrt{100c^2 – 36c^2}}{6} = \frac{10c \pm 8c}{6} = \frac{5\pm 4}{3}c$, so $b = \pm\sqrt{\frac{5\pm 4}{3}}a$.

This is real, then, when $x = a \pm\sqrt{\frac{5\pm 4}{3}}ai$.

This is imaginary when $a^6 – 15a^4b^2 + 15a^2b^4 – b^6 = c^3 – 15c^2d + 15cd^2 – d^3 = 0$.

We can factor $c^3 – 15c^2d + 15cd^2 – d^3 = (c – d)(c^2 – 14cd + d^2)$. $c – d = 0$ when $c = d$, so $b = \pm a$ is one possibility. Let’s look at the other case.

$c^2 – 14cd + d^2 = 0$ when $d = \frac{14c \pm\sqrt{196c^2 – 4c^2}}{2} = 7c \pm\sqrt{48}c = (7 \pm 4\sqrt{3})c$, so $b = \pm\sqrt{7 \pm 4\sqrt{3}}a$.

This is imaginary, then, when $x \in \{a \pm ai, a \pm\sqrt{7\pm 4\sqrt{3}}ai\}$.

Summary

At $k = 7$, we have the cubics $c^3 – 21c^2d + 35cd^2 – 7d^3$ and $7c^3 – 35c^2d + 21cd^2 – d^3$, which do not have factors, so this is a good place to stop.

Factoring out $a$, here are the cases where $x\in\mathbb{C}^*$ yields $x^k \in \mathbb{R} \lor \mathbb{I}$:

k	$x^k\in\mathbb{R}$	$x^k\in\mathbb{I}$
1	(None)	(None)
2	(None)	$1 \pm i$
3	$1\pm\sqrt{3}i$	$1\pm\frac{\sqrt{3}}{3}i$
4	$1\pm i$	$1 \pm(3 \pm 2\sqrt{2})i$
5	$1\pm\sqrt{5\pm 2\sqrt{5}}i$	$1\pm\sqrt{1\pm\frac{2\sqrt{5}}{5}}i$
6	$1\pm\sqrt{\frac{5\pm 4}{3}}i$	$\{1\pm i, 1\pm\sqrt{7\pm 4\sqrt{3}}i\}$